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Simultaneous equations.

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    160 cos(theta) - W cos(69) = 0

    160 sin(theta) - W cos(21) - W = 0

    Can someone solve these equations for me to find W and theta?
     
  2. jcsd
  3. Sep 22, 2009 #2

    rock.freak667

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    160cosθ=Wcos(69)

    W=160cosθ/cos(69)


    continue now
     
  4. Sep 22, 2009 #3
    W = W cos(69) / cos(69)


    ?
     
  5. Sep 22, 2009 #4

    rock.freak667

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    no, put W=160cosθ/cos(69) into the second equation
     
  6. Sep 22, 2009 #5
    Sorry I am still not seeing it. Could you give me one more step?
     
  7. Sep 22, 2009 #6

    rock.freak667

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    In this equation

    160 sinθ - W cos(21) - W = 0


    everywhere you see 'W' put 160cosθ/cos(69) and try to simplify it.
     
  8. Sep 22, 2009 #7
    I do understand how to simplify one equation and substitute into the other, however I just can't seem to simplify these down. The theta with the sin and cos attached is what is really confusing me when trying to simplify. This is just a mess and I'm very confused.
     
  9. Sep 22, 2009 #8

    rock.freak667

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    just show where you've reached and we'll see if we can simplify it further
     
  10. Sep 22, 2009 #9
    160 sinθ - (160cosθ/cos69) cos(21) - (160cosθ/cos69) = 0
     
  11. Sep 22, 2009 #10

    rock.freak667

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    so you get


    [tex]160 sin \theta - (\frac{160cos21}{cos69} - \frac{160}{cos69})cos \theta = 0[/tex]


    this is like Asinθ-Bcosθ=0, how do you solve this equation?
     
  12. Sep 22, 2009 #11
    Is there a trig identity that replaces either sin or cos? I'm thinking we want to narrow this down to only having either sin or cos in the equation.
     
  13. Sep 22, 2009 #12

    rock.freak667

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    well yes but we can put it like this


    Asinθ=Bcosθ

    or sinθ/cosθ=B/A

    do we know another way to express sinθ/cosθ ?
     
  14. Sep 22, 2009 #13
    I am really not sure.
     
  15. Sep 22, 2009 #14

    rock.freak667

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    do you not know that tanθ=sinθ/cosθ ?
     
  16. Sep 22, 2009 #15
    Oh my it has been a long day.. Yes i see that now. So simplifying I would then do:

    160 sinθ - (-29.6542) cosθ = 0

    160 sinθ = 29.6542 cosθ

    θ = tan^-1(160/29.6542)

    θ = 79.5° ?
     
  17. Sep 22, 2009 #16

    rock.freak667

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    yes that should be correct. You are able to find W knowing θ=79.5° right?
     
  18. Sep 22, 2009 #17
    Yes:

    160cos(79.5) - W cos(69) = 0

    29.1577 = W cos(69)

    W = 81.36
     
  19. Sep 22, 2009 #18
    Thank you so much for all your help!
     
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